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Simplify (2t^–3)^3(0.4r)^2 Checking answers I got 1.28r^2 over t ^-6
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You have:\[(2t^{–3})^3(0.4r)^2 \]So this is:\[2^3(t^{-3})^30.4^2r^2\] Of course, 2³=8 and (0.4)²=0.16. 8*0.16=1.28, so that's fine. I see two problems in your answer: 1. t^-6 remember: (a^b)^c=a^(b*c), so... 2. "over" if you want to make a fraction with t in the denominator, you'll lose the - of the exponent.
thanks
YW! I got: \[1.28t^{-9}r^2=\frac{ 1.28r^2 }{ t^9 }\]
Writing a decimal fraction within a proper fraction looks ugly, so It would be better to write 1.28 as 128/100=32/25, so the final answer would be:\[\frac{ 32r^2 }{ 25t^9 }\]
that was right
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